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emcrazy14
 2 years ago
Differentiate;
x^2/(1+x)^2
emcrazy14
 2 years ago
Differentiate; x^2/(1+x)^2

This Question is Closed

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.2This is of the form u/v. (u/v)'=(u'vv'u)/v^2 Here u=x^2 and v=(1+x)^2. (x^n)'=nx^(n1) Can u do it @emcrazy14?

emcrazy14
 2 years ago
Best ResponseYou've already chosen the best response.0I got the answer 2x/(1+x)^3. Is it correct?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1differentiate: 1/(1/x + 1)^2

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.2check the sign @emcrazy14. It shud be 2x/(1+x)^3

emcrazy14
 2 years ago
Best ResponseYou've already chosen the best response.0Ok, thanks. Do I multiply (x+1) with a negative sign to change it into (1+x) so that it cancels with the the denominator (1+x)^4?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1just ask the wolf for differentiation Question.

emcrazy14
 2 years ago
Best ResponseYou've already chosen the best response.0Then how oes it become +2x? :O

emcrazy14
 2 years ago
Best ResponseYou've already chosen the best response.0@experimentX : wolf? :O *does

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=differentiate+x^2%2F%281%2Bx%29^2

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.2dw:1352469822062:dwdw:1352469940166:dwdw:1352470079425:dw

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1\[ x^2/(1+x)^2 = 1/(1/x + 1)^2 = ({1 \over x} + 1)^{2}\] use chain rule ... this should be shorter.

emcrazy14
 2 years ago
Best ResponseYou've already chosen the best response.0@ajprincess : Get it now. Thank you so much! :D

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.2welcome:) @experimentX's method is much more simpler. U can use that too.:)

emcrazy14
 2 years ago
Best ResponseYou've already chosen the best response.0@experimentX : Once I perfect this one, I'll use the one you mentioned. Thank you! :)
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